Why Disjoint Gaussian Moving-Average Outputs Are Independent

# Show that $Y_t$ and $Y_{t+h}$ are independent if $X_t$ is Gaussian


# Show that $Y_t$ and $Y_{t+h}$ are independent if $X_t$ is Gaussian












If $Y_t=\sum_{i=0}^qa_iX_{t-i}$ where $X_{t-i}$ is Gaussian with mean $\mu$ and variance $\sigma^2$, how do I show that $Y_t$ and $Y_{t+h}$ are independent (for $|h|>q$) using the joint pdf. I know this question has been asked before (here) but the answer is not what I'm looking for since independence is not shown, but only that $Y_t$ is Gaussian. I want to show independence between $Y_t$ and $Y_{t+h}$ by writing their joint pdf as the product of their marginal pdf's.

## Answer by mark leeds (score 1)

https://quant.stackexchange.com/a/63511

Hi Parseval: Let me put an answer here and hopefully it's clear. I fixed it so that $\mu = 0 $ is not needed but of course, the independent Gaussian assumption is still needed.

(1) $Y_t = \sum_{i=0}^{q} a_{i} X_{t-i}$

(2) $Y_{t+h} = \sum_{j=0}^{q} a_j X_{t+h-j}$

$ h > q$.

Note that the last (earliest ) term in (2) is $a_{q} X_{t+h-q}$ and first (latest ) term in (1) is $a_{0} X_{t}$. Therefore, since $h > q$, none of the noise terms in $Y_t$ overlap with any of the the noise terms in $Y_{t+h}$.

Now, the usual definition of covariance, gives:

$Cov(Y_{t}, Y_{t+h}) = E(\sum_{i=0}^{q} a_{i} X_{t-i} \sum_{j=0}^{q} a_{j} X_{t+h-j}) - E(\sum_{i=0}^{q} a_{i} X_{t-i}) E(\sum_{j=0}^{q} a_j X_{t+h-j}) $

So, for the first term we have the two sums multiplying each other and then an expectation is taken. Then, for the second term, we have two expectations multiplying each other.

FIRST SIMPLIFY THE SECOND TERM

Taking the expectations of the terms, we get $\sum_{i=0}^{q} a_{i} \mu \sum_{j=0}^{q} a_j \mu = \mu^2 \sum_{i=0}^{q} a_{i} \sum_{j=0}^{q} a_j$.

Simplifying this, results in:

$ \mu^2 \sum_{i=0}^{q} \sum_{j=0}^{q} a_i a_j $

NOW SIMPLIFY THE FIRST TERM:

$ E(\sum_{i=0}^{q} a_i X_{t-i} \sum_{j=0}^{q} a_j X_{t+h-j}) = $

$( \sum_{i=0}^{q} \sum_{j=0}^{q} a_{i} a_j ) E(X_{t-i} X_{t+h-j})$

But we showed earlier than that none of the terms in the very last expectation overlap, and, since they are independent Gaussians, we can re-write the last expression as $ E(X_{t-i} X_{(t+h-j)}) = E(X_{t-i}) E(X_{(t+h-j)}) = \mu^2$.

So, we showed that the first term equals the same thing as the second term which means that $Cov(Y_{t}, Y_{t+h}) = 0 $.

Note though that I needed the assumption that the X_{t} are independent Gaussian RV's but this is the usual assumption in the MA(q) model.

## Answer by steveo'america (score 0)

https://quant.stackexchange.com/a/63523

If you are going to assume the $X_t$ are independent, then write $\vec{X}$ as the vector of $X_{t-q},\ldots,X_t,X_{t+h-q},\ldots,X_{t+h}$. Then $\vec{X} \sim \mathcal{N}\left(\mu \vec{1},\Sigma\right)$, where $\Sigma$ is a diagonal matrix. Then the vector of the two $Y$ values is $A \vec{X}$ for some block matrix that has one row of $q+1$ of the $a_i$, then $q+1$ zeros, then $q+1$ zeros and then the $a_i$ in the second row. Simple to confirm that $A\Sigma A^{\top}$ is diagonal.